Carryovers occur at pairs of matching numbers between two rows, and the carryover is the number decrement by one

@TedShifrin The Hodge star $\star$ on $\Lambda^2\Bbb R^4$ has $\pm1$ eigenspaces $\Lambda^2_L$ and $\Lambda^2_R$ of left/right isoclinic rotations. This can be extended to an antilinear operator on $\Lambda^2\Bbb C^4$. Then ${\rm SL}(4,\Bbb C)$ respects the symmetric bilinear form $$\wedge:\Lambda^2\Bbb C^4\times\Lambda^2\Bbb C^4\to\Lambda^4\Bbb C^4$$ and ${\rm U}(4)$ commutes with $\star$.

So ${\rm SU}(4)$ acts on the new $\pm1$ eigenspaces, which are $\Lambda^2_L+i\Lambda_R^2$ and $\Lambda_R^2+i\Lambda_L^2$, each of which has $6$ real dimensions and which has $\wedge$ as a definite real inner product. This gives the double covering ${\rm SU}(4)\to{\rm SO}(6)$.

If the pi position of ones sequence has a closed form, then it is easier to analyse where and how the carryovers will occur

At least early on, it seems that there are many ones in pi that occupy positions that are multiples of 3

Meanwhile the transcendental number $a$ has positions given by the triangular numbers

The triangular numbers are multiples of 3 when:

@Danu Mike Miller's idea makes sense to me: $D\ominus\ell$ being the relative orthogonal complement. (Since $D$ is complex 2D and $\ell$ is complex 1D, it should also be a line.) The fiber of $\nu\in\ell^a$ should be all pairs $(\ell,v\oplus\ell)$? I haven't checked any of this to see if it makes sense or read the transcript further.

I'll have to think about ${\rm SO}(6)/{\rm U}(3)\cong {\rm SU}(4)/{\rm S}({\rm U}(3)\times{\rm U}(1))$ later.

$n(n+1)=6m\implies n^2+n-6m=0\implies n=\frac{-1\pm \sqrt{1+6m}}{2}$

which has integer solutions $n$ when 1+6m is a perfect square

Likewise, the triangular numbers that are multiples of 6 requires 1+12m to be a perfect square

@LeakyNun yelloha

@KasmirKhaan hi

My teacher told me to put all the exercices on one pdf

can you please help me with that?

maybe

Therefore, given a product of transcendentals, ab, it might be a fruitful question to query the theorems what the closed form of the position of ones are in its binary expansion for a and b, which then by investigating the nature of the integers involved (squarefree, prime, square, multiplies etc.) will give exact information on which positions have no first order carryovers

@LeakyNun is that a polite way to say no ? ><

@KasmirKhaan no it isn't

@LeakyNun Ehm btw , still working on those exercices you gave me yesterday =p think be done with em in few hours , had to do other stuff

@KasmirKhaan alright

:D

I won't be here often in the following three days

oh :(

why ?

@Secret have you watched the video?

@KasmirKhaan camp

@LeakyNun oh nice :D hope you lots of fun :)

If a closed form cannot be found, either provably or due to physical limitations, we can instead use computers to generate a huge number of digits, and then formulate the questions in terms of bayesian statistics (e.g. asking what is the probability of the positions of ones in the binary expansion of $\pi$ are multiples of 3). Then within some error limits, we can say how irrational a given product of transcendental is

@LeakyNun I have watched the bits that talked about the powers of the liuoville's number

In programming you have lists or arrays. What is the mathematical equivalent for that? A vector?

@Kasper a tuple

(the programming language R refers to them as vector, which is highly inaccurate)

Python also has tuples

but the n-tuple (x_1,x_2,...,x_n) is kind of like the vector of dimension n right?

@Kasper yes, kind of like.

don't get me started :P

@Kasper Except mathematically, a vector is an element of a vector space, which is not quite the same

yeah okay, so matrix nxn multplied by n-tuple is not defined

but matrix times n-vector is

can I think of it like that?

you can.

that depends on your background.

@Secret :o this is very amazing:

consider the binary expansion of the real numbers from 0 to 1

we define a relation $\sim$ as follows:

$a \sim b$ iff $a$ and $b$ differ at finitely many bits

we see that it is actually an equivalence relation:

1. reflexive: any number differs with itself at exactly 0 bits, where 0 is a finite number

2. symmetric: come on

3. transitive: well, that needs more pondering but is still trivial

so we can partition $[0,1]$ using $\sim$...

might be useful if some given transcendental number is defined via some kind of bounds similar to the Liouville numbers, because the equvalence relation will mean there exists some sums of Liouville numbers that are also liouville numbers, but that's just a guess...

and we knew that all liouville numbers are transcendental, except I don't quite understood what is $A(\alpha)$ doing in the proof

@Secret I think it just means that $A$ depends on $\alpha$.

Oh and if $a\sim b$ it means |a-b| is a rational that is a finite sum of dynadic fractions (because all finite sums of dynadic fractions can be represented using finite number of bits)

@LeakyNun But why we need $A$?

@Secret hmm, interesting

so it's actually just $[0,1]/\Bbb Q_2$ where $\Bbb Q_2$ denotes the dyadic rationals

@Secret $\alpha$ is irrational algebraic $\implies$ A exists and (6) holds. However, since $\alpha$ is Liouville, (6) cannot hold, so $\alpha$ is transcendental.

ok

It's a good thing that the lab day on the first week of class is pretty much a meet-and-greet

like computing lab?

Otherwise I'd be furious with the absent lab TA I'm having to cover for, rather than merely peeved

Intro physics lab

I really should be doing chemistry, but I fell into the mathematics black hole

I get to campus at the start of the day, because of my

I read that a $n \times n$ matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors. Does it matter to which eigenvalues they linearly independent eigenvectors are associated? For instance, could all $n$ linearly independent eigenvectors be associated with a single eigenvalue?

I get to campus at the start of the day, and noticed some student confusion re where they were supposed to be

@user193319 Eigenvectors associated to different eigenvalues are always linearly independent

Eigenvectors corresponding to distinct eigenvalus are linearly independent

They got routed eventually but their TA hadn't arrived yet so I was asked to keep an eye on their students

But eigenvectors associated to a same eigenvalues are not necessarily linearly independent

That was 45 minutes ago...

hey! any help on Thomas algoriithm to solve for tridiagonal system

And, yes, a matrix can have all linearly independent eigenvectors with the same eigenvalue

e.g. zero matrix, unit matrix, ...

eigenvalue, woops

last part "same eigenvalue" perhaps

yes

while solving ODE's through Finite difference scheme I am getting to tridiagonal system through Thmas method

which is a bit confusing to me

Okay. I see. Thanks everyone.

Any multiple of the identity matrix, really

ams.org/journals/notices/201606/rnoti-p604.pdf

Seems eigenvalues also works for tensors

Not so surprising, really. If you've got a nD array but you specify all but two of the indices, then these two that remain comprise a matrix

I wonder, if a tensor not only have eigenvectors, but also eigensubtensors...

@anon Maybe I'm being stupid, but I don't see how this works since $\ell$ and $D$ are isotropic, in particular the orthogonal complement of $\ell$ is just $D$ itself

@anon OK. Thanks for your time and I'll be looking forward to any further comments ;D

If a matrix $A \in \mathbb{R}^{n \times n}$ has $n+1$ eigenvectors, such that every $n$ of them are linearly independent, is $A$ a scalar matrix?

I think so...

Ah yeah, I got it. Solution: Write one of them as linear combination of the others, see that all coefficients are nonzero by the imposed condition and then get a contradiction to the fact that this vector is also an eigenvector.

If you name the first $n$ eigenvectors $e_1$ to $e_n$ and their eigenvalues $\lambda_1$ to $\lambda_n$, then after a change of basis $A$ must be the diagonal matrix with $\lambda_1, \dots, \lambda_n$ on the diagonal

And if you name the final one $e_{n+1}$ with eigenvalue $\lambda_{n+1}$, then it is also possible to diagonalise $A$ as the diagonal matrix with $\lambda_1, \dots, \lambda_{n-1},\lambda_{n+1}$ on the diagonal

But then $\lambda_n = \lambda_{n+1}$

And you can do the same for any other eigenvalue, so they're all the same

So after diagonalisation $A$ is scalar. But since scalar matrices commute with all other matrices, $A$ must equal this scalar matrix

That would be my approach - no guarantees on it being correct, though :P

hi yall

if 0<=i <n and 0<=j<n , how can we bound i-j

-n < i-j < n

$0\le i<n$ and $0\le j<n$?

@LeakyNun hmm how did you get that bound , like algebraicly ?

and cant we do better?

$0\le i<n$ and $-n<-j\le 0$

@AkivaWeinberger yes sir

@KasmirKhaan ^

@AkivaWeinberger oh that was the answer >< thought u written it in latex language only ><

Adding, those two, we get $-n<i-j<n$. (Why does adding $\le$ and $<$ yield $<$?)

Which gives us Leaky's answer.

chat.SE is responding slower than usual for me

@LeakyNun must be your internet connection , happens to me also from time to time

@KasmirKhaan other pages are fine for me

@LeakyNun hmm try to restart it

@KasmirKhaan it seems to be working fine now

:D

am trying to prove btw why the elements <a> are only {e,a,a^2 ....a^n-1}

@KasmirKhaan what is the definition of <a>?

so i have to show that if we pick any two different elements in there, they are different

generator

and second part that any other power is in <a>

$n$ is the order of $a$?

chat is now being slower again

@KasmirKhaan of course I know what it is. I'm asking you what it is.

I can follow thru now knowing that inquility

oh I thought it was different notation ><

ehm its the powers of that element

generated by a

@AkivaWeinberger order of a is n

How are you defining the order

the power least non zero number such that we rise it to a we get e

@AkivaWeinberger

@AkivaWeinberger still trying to figure out why we deleted the equal signs of that enquialiy =p

@KasmirKhaan Suppose $a<b$ and $c\le d$. We want to show that $a+c=b+d$ is impossible.

Either $c<d$, in which case adding $a<b$ and $c<d$ gives $a+c<b+d$,

@AkivaWeinberger they were both less than or equal

I think I wrote it wrong

@KasmirKhaan it's <n.

yes

Oh, both? So this:

was incorrect?

Yes

><

that was correct

less than or equal only on the zero part

Oh

both stricktly less than n

@KasmirKhaan it should be <n.

Oh, OK, so

chat.SE is behaving badly.

the point is $i-j$ can't be exactly $n$

@LeakyNun yes but he said sometthing about removing the equal sign i wanted to understand

This MO question has a deep connection with proving RH and IMO 2017 P6 https://mathoverflow.net/questions/280429/bound-on-number-of-nxn-grids-with-lex‌​icographical-ordering-poset-structure

Yes exactly @AkivaWeinberger

If we want $i-j$ to be as large as possible, then we want $i$ to be as large as possible and $j$ to be as small as possible

That would give us $n-0$, except that $i$ can't exactly equal $n$

Oh yeah yeah =p

so $i-j$ would be less than $n$

I get it now

j can be 0

and i can be n-1

so it is less than n for sure =p

okay all clear now :D finishing the proof :D

@LeakyNun @AkivaWeinberger thanks ! :D

In any case, yeah, you want to show that any two of the things in that set are different, and that $a^k$ is in that set for every integer $k$

Anybody checked the MO question I linked above ^^ ?

...dammit

Why ?

You know why

Very brilliant connection, isn't it ?

what is that link

@KasmirKhaan Check a bit above chat.stackexchange.com/transcript/message/39829435#39829435

I never open links from guys named Alex :D

Am scared is it a virus?

-.-

@KasmirKhaan Akiva opened it. And yeah, it's a ransomware. Isn't it, @AkivaWeinberger ?

$\langle a,b \mid a^2b^{-2} \rangle$

@AkivaWeinberger what is this group?

@AlexKChen goo.gl/eZS7ga

@LeakyNun Oh, I remember dealing with that group a while back

I think people were asking if there were any elements of order $2$ in it (there aren't)

It's also not abelian

The Cayley graph looks interesting

I don't know much more about it than what can be learned from the Cayley graph, though

I think it's the fundamental group of the Klein bottle.

So in particular a group of isometries of $\Bbb R^2$.

(In particular it can be written as a semidirect product of $\Bbb Z$ with itself by $\text{-id}$ map)

@BalarkaSen Oh, right, yeah

Forgot about that

Mm. That makes sense. If $a$ is a glide reflection and $b$ is the corresponding translation, then $a^2$ definitely equals $b^2$

right

Wait hold on that's not how the Klein bottle stuff works

Is it?

? You identify one side of the square to the opposite by a glide reflection

Isn't the fundamental group of the Klein bottle $\langle a,b|abab^{-1}\rangle$? Or are those the same?

Yeah those are the same. abab^-1 = 1 is the same as (ab)^2 = b^2

Isomorph that to <x, y|x^2 = y^2>

Ah, I see

hi chat

hello

hey

Yo

@Daminark did you see my message?

I'll check

LOL

hmm. if $y=1008$ mod 2011, then $2y=2016=5$

and then $402(2y)=5(402)=2010=-1$.

so $-804y=(2011-804)y=1207y=1\implies y=1008=1207^{-1}$...going nowhere fast

standard modpow is O(n) time

I mean, since $p=2011$ is prime you can say that $a^p=a$ mod $p$

which effectively means that you can take the exponent mod p-1

that gets you down to $1008^{702}$.

@anon I guess maybe with respect to the Hermitian scalar product we can take the orthogonal complement... Am I making any sense? :P

@Semiclassical yes

how to do better ?